Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in PYTHON):

Solution:
n=int(raw_input())
x=map(int,raw_input().split())
y=map(int,raw_input().split())

cnt=0
ans1=[]
ans2=[]
bit=dict()
for i in range(30,-1,-1):
    for j in range(0,n):
        if x[j]&1<<i > 0 and x[j] < 1<<i+1:
            if i not in bit:
                bit[i] = j
            else:
                x[j] ^= x[bit[i]]
                ans1.append((j,bit[i]))
can = 1
for i in range(30,-1,-1):
    has = 0 
    for j in range(0,n):
        if y[j]&1<<i > 0 and y[j] < 1<<i+1:
            has = 1
            if i not in bit:
                can = 0
            else:
                if j == bit[i]:
                    continue
                elif y[bit[i]]&1<<i == 0:
                    y[j],y[bit[i]] = y[bit[i]],y[j]
                    ans2.append((j,bit[i]))
                    ans2.append((bit[i],j))
                    ans2.append((j,bit[i]))
                else:
                    y[j] ^= y[bit[i]]
                    ans2.append((j,bit[i]))
    if not can:
        break
    if has:
        for k in range(i,-1,-1):
            p1 = x[bit[i]]&1<<k
            p2 = y[bit[i]]&1<<k
            if p1 == p2:
                continue
            else:
                if k not in bit:
                    can = 0;
                    break
                else:
                    x[bit[i]] ^= x[bit[k]]
                    ans1.append((bit[i],bit[k]))
if not can:
    print -1
else:
    for j in range(0,n):
        if x[j] != y[j]:
            ans1.append((j,j))
    print len(ans1)+len(ans2)
    for j in range(0, len(ans1)):
        print ans1[j][0]+1,ans1[j][1]+1
    for j in range(len(ans2)-1, -1, -1):
        print ans2[j][0]+1,ans2[j][1]+1